Reply To: Challenge Questions – 2

Home Forums VMTTC Online Discussion Forum Challenge Questions – 2 Reply To: Challenge Questions – 2

#33105
Preeti Pathak
Participant

Hi,

challenge question #21

since in a triple, one  element is always divisible by 3.That means , if we consider one of the elements, for example ,to be either

3 / 6 /9 / 12 / 15 / 18 / 21 /24/27/30  etc. (multiples of 3).Their corresponding squares will be,

9/36/81/144/324/441/729/900.And their corresponding digit sums will be

9/9/9/9/9/9/9 .And 9 ‘s can be casted out.

so the digit sum of squares of one element(other than the one divisible by 3) will be equal to the digit sum of square of the 3rd element(of course, the remaining element).

we can verify this by applying the digit sum check.

For example

  1.  TRIPLET  3, 4 5———————-digit sum check—————->3 IS DIVIBLE BY 3 ,SO RULED OUT.
  2.                   L.H.S=  DS(4^2 )=DS(16)=7,R.H.S=DS(5^2)= DS(5^2)=DS(25)=7 ,L.H.S=R.H.S

2.TRIPLET 24, 7 25——————–digit sum check——————->24 IS DIVIBLE BY 3 ,SO RULED OUT.

L.H.S=DS(7^2)=DS(49)=4,R.H.S=DS(25^2)DS(625)=4,L.H.S=R.H.S

Thanks

error:
Scroll to Top
Visit Us On FacebookVisit Us On Youtube